Question: Divide the following complex numbers. $ \dfrac{1-5i}{1+i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1-i}$ $ \dfrac{1-5i}{1+i} = \dfrac{1-5i}{1+i} \cdot \dfrac{{1-i}}{{1-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(1-5i) \cdot (1-i)} {(1+i) \cdot (1-i)} = \dfrac{(1-5i) \cdot (1-i)} {1^2 - (1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(1-5i) \cdot (1-i)} {(1)^2 - (1i)^2} = $ $ \dfrac{(1-5i) \cdot (1-i)} {1 + 1} = $ $ \dfrac{(1-5i) \cdot (1-i)} {2} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({1-5i}) \cdot ({1-i})} {2} = $ $ \dfrac{{1} \cdot {1} + {-5} \cdot {1 i} + {1} \cdot {-1 i} + {-5} \cdot {-1 i^2}} {2} $ Evaluate each product of two numbers. $ \dfrac{1 - 5i - 1i + 5 i^2} {2} $ Finally, simplify the fraction. $ \dfrac{1 - 5i - 1i - 5} {2} = \dfrac{-4 - 6i} {2} = -2-3i $